Question

Does Aspirin Prevent Heart Disease? In a trial designed to test the effectiveness of aspirin in preventing heart disease, 11,037 male physicians were treated with aspirin and 11,034 male physicians were given placebos. Among the subjects in the aspirin treatment group, 139 experienced myocardial infarctions (heart attacks). Among the subjects given placebos, 239 experienced myocardial infarctions (based on data from “Final Report on the Aspirin Component of the Ongoing Physicians’ Health Study,” New England Journal of Medicine, Vol. 321: 129–135). Use a 0.05 significance level to test the claim that aspirin has no effect on myocardial infarctions.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. Based on the results, does aspirin appear to be effective?

Short answer

a. Using the hypothesis test, it is concluded thatthere is sufficient evidence to reject the claim thataspirin has no effect on myocardial infarctions.

b.Thus, the 95% confidence interval is equal to (-0.12, -0.006). As 0 does not lie within the interval, there is sufficient evidence to reject the claim thataspirin has no effect on myocardial infarctions.

c. It can be said that aspirin is effective in reducing the risk of myocardial infarction.

Step-by-step solution

Given information

A sample of 11,037 male physicians was treated with aspirin. Among them, 139 experienced myocardial infarctions. Another sample of 11,034 male physicians was given placebos. Among them, 239 experienced myocardial infarctions.

Describe the hypotheses

It is claimed that aspirin has no effect on myocardial infarctions.

Null hypothesis: Theproportion of physicians who experienced myocardial infarctions after taking aspirin is equal to the proportion of physicians who experienced myocardial infarctions after taking the placebo.

$H_0:p_1=p_2$

Alternative hypothesis: Theproportion of physicians who experienced myocardial infarctions after taking aspirin is not equal to the proportion of physicians who experienced myocardial infarctions after taking the placebo.

$H_1:p_1\ne p_2$

The test is two-tailed.

Important values

Let ${\hat{p}}_1$denote the sampleproportionof physicians who experienced myocardial infarctions after taking aspirin.

$$\begin{gathered} {\hat{p}}_1=\frac{139}{11037} \\ =0.0126 \end{gathered}$$

Let ${\hat{p}}_2$denote the sample proportion of physicians who experienced myocardial infarctions after taking the placebo.

$$\begin{gathered} {\hat{p}}_2=\frac{239}{11034} \\ =0.0217 \end{gathered}$$

The sample size of physicians who weretreated with aspirin$\left(n_1\right)$ is equal to 11037.

The sample size of physicians who weretreated with a placebo$\left(n_2\right)$ is equal to 11034.

The value of the pooled sample proportion is computed as follows.

$$\begin{gathered} \overline{p}=\frac{\left(x_1+x_2\right)}{\left(n_1+n_2\right)} \\ =\frac{\left(139+239\right)}{\left(11037+11034\right)} \\ =0.0171 \end{gathered}$$

.$$\begin{gathered} \overline{q}=1-\overline{p} \\ =1-0.0171 \\ =0.9829 \end{gathered}$$

Find the test statistic

The test statistic is computed as follows.

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\left(\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}\right)}} \\ =\frac{\left(0.0126-0.0217\right)-0}{\sqrt{\left(\frac{0.0171\times 0.9829}{11037}+\frac{0.0171\times 0.9829}{11034}\right)}} \\ =-5.1907 \end{gathered}$$

The value of the test statistic is -5.1907.

Referring to the standard normal distribution table, the critical value of z corresponding to $\alpha =0.05$for a two-tailed test is equal to 1.96.

Referring to the standard normal distribution table, the corresponding p-value for z equal to -5.1907 is equal to 0.000.

As the p-value is less than 0.05, the null hypothesis is rejected.

Conclusion of the test

a.

There is sufficient evidence to reject the claim thataspirin has no effect on myocardial infarctions.

Find the confidence interval

b.

The general formula for the confidence interval of the difference of proportions is written below.

$\text{Confidence} \text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E , \left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) ...\left(1\right)$

Here, E is the margin of error and has the following formula:

$E=z_{\frac{\alpha }{2}}\times \sqrt{\left(\frac{{\hat{p}}_1\times {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\times {\hat{q}}_2}{n_2}\right)}$

For computing the confidence interval, first, find the critical value$z_{\frac{\alpha }{2}}$.

Here, $\alpha =0.05$.

Hence,

$$\begin{gathered} \frac{\alpha }{2}=\frac{0.05}{2} \\ =0.025 \end{gathered}$$

The value of $z_{\frac{\alpha }{2}}$from the standard normal table is equal to 1.96.

Now, the margin of error (E) is equal to

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\left(\frac{{\hat{p}}_1\times {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\times {\hat{q}}_2}{n_2}\right)} \\ =1.96\times \sqrt{\left(\frac{0.0126\times 0.9874}{11037}+\frac{0.0217\times 0.9783}{11034}\right)} \\ =0.00342 \end{gathered}$$

.

Substitute the value of E in equation (1), as follows.

.$$\begin{gathered} \text{Confidence} \text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E , \left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) \\ =\left(\left(0.0126-0.0217\right)-0.00342 , \left(0.0126-0.0217\right)+0.00342\right) \\ =\left(-0.012 , -0.006\right) \end{gathered}$$

Thus, the 95% confidence interval is equal to (-0.12, -0.006).

Conclusion based on the confidence interval

The interval does not contain the value of 0 and contains all negative values.

This implies that the values of the two proportions cannot be equal.

There is sufficient evidence to reject the claim thataspirin has no effect on myocardial infarctions.

Effectiveness of aspirin

c.

As the confidence interval consists of negative values, the proportion of physicians who experienced myocardial infarctions after taking aspirin is less as compared to the physicians who took the placebo.

Therefore, it can be concluded that aspirin is effective in reducing the case of myocardial infarction.